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@ Appl. Gen. Topol. 19, no. 2 (2018), 245-252doi:10.4995/agt.2018.8981
c© AGT, UPV, 2018

The hull orthogonal of the unit interval [0,1]

Sami Lazaar
a
and Saber Nacib

b

a
Faculty of Sciences of Tunis, University of Tunis El Manar, Tunisia. (salazaar72@yahoo.fr)

b
Faculty of Sciences of Gafsa, University of Gafsa, Tunisia. (sabernacib83@hotmail.com)

Communicated by S. Romaguera

Abstract

In this paper, the full subcategory Hcomp of Top whose objects are

Hausdorff compact spaces is identified as the orthogonal hull of the unit

interval I = [0, 1]. The family of continuous maps rendered invertible
by the reflector β ◦ ρ is deduced.

2010 MSC: 54D30;18B30; 54D60.

Keywords: completely regular spaces; categories; Stone-Čech compactifica-

tion.

1. Introduction

In the literature, various approaches to the Stone-Čech Compactification βX
of a topological space X are given, using constructions based on products of
the interval unit I, ultrafilters, and C⋆-algebras, respectively ([5], [7], [12] and
[10]).

More than a compactification, the embedding of X into βX defines Hcomp
as a reflective subcategory in the category Tych of Tychonoff spaces. Thus
Hcomp is a reflective subcategory of Top with reflector β ◦ ρ, where ρ is the
Tychonoff reflector.

The year 1937 was an important one in establishing nice connections between
topology and algebra. M. H. Stone and E. Čech published papers giving several
fundamental properties of the compactification βX, which had been introduced
by Tychonoff. For instance, Stone showed that any Tychonoff space X is C⋆-
embedded in βX, and this can be interpreted algebraically as showing that the
rings C⋆(X) and C⋆(βX) are isomorphic.

Received 22 November 2017 – Accepted 06 May 2018

http://dx.doi.org/10.4995/agt.2018.8981


S. Lazaar and S. Nacib

Recall that if D is a reflective subcategory in a category C, with reflector
F, then D⊥ = {f ∈ homC : F(f) is an isomorphism} and D

⊥⊥ = D (for

more information see [1], [2] and [4]). In our case, we have Hcomp⊥ = {f ∈

homTop : β ◦ ρ(f) is an isomorphism} and Hcomp
⊥⊥ = Hcomp.

So on the one hand, if we consider the category Sob of sober spaces, it is
not difficult to show that Sob⊥ = {δ}⊥, where δ is the Sierpiński space, and
thus Sob = {δ}⊥⊥ which gives a characterization of sober spaces using only
the space δ.

On the other hand, in [6], A. Haouati and S. Lazaar showed that the reflective
subcategory Hewitt of Top, whose objects are real-compact spaces, is the
orthogonal hull of the real line R.

Analogous to Sob⊥ = {δ}⊥ and Hewitt⊥ = {R}⊥, we show in this paper

that Hcomp⊥ = {I}⊥ where I is the unit interval, and consequently the family
of continuous maps rendered invertible by β ◦ ρ are those maps which are
orthogonal to I.

2. Some preliminary results

Let C be a category. An arrow f in C from A to B is said to be orthogonal
to an object X in C if and only if for any arrow g from A to X, there exists a
unique arrow g̃ from B to X satisfying g̃ ◦ f = g.

The orthogonal Σ⊥ of a class of morphisms Σ is the class of objects orthog-
onal to every morphism in Σ [4]. The orthogonal of a class of objects is defined
analogously.

Recall that a topological space is called completely regular (or Tychonoff) if
it is T1 and every closed subset F of the space is completely separated from any
point x not in F . An other important characterization of completely regular
spaces is given by the following theorem.

Theorem 2.1 ([12, Proposition 1.7]). A space is completely regular if and
only if the family of zero-sets of the space is a base for the closed sets (or
equivalently, the family of cozero-sets is a base for the open sets).

Notations 1. Lat X be a topological space. We denote by:

• C(X) the family off all continuous maps from X to R.
• C⋆(X) the family off all bounded continuous maps from X to R.
• C⋆I(X) the family off all continuous maps from X to I.
• C[0,+∞[(X) the family off all positive continuous maps from X to R.

Remark 2.2. Let f be a continuous map from a topological space X to R.
Consider the map fI from X to I defined by fI := inf{|f|, 1}.

Clearly f(x) = 0 if and only if fI(x) = 0 if and only if |f|(x) = 0. Therefore
by Theorem 2.1 a topological space X is completely regular if and only if the
family {h−1(]0, 1]) : h ∈ C⋆I(X)} (resp., {h

−1(]0, +∞[) : h ∈ C[0,+∞[(X)} is a
base for the open sets of X.

The following result is an easy observation from [2].

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The hull orthogonal of the unit interval [0,1]

Proposition 2.3. Let C be a category and D a reflective subcategory of C, with
reflector F. An arrow in C is orthogonal to an object in D if and only if its
F-identification is also.

Remark 2.4. In our case when we consider the reflective subcategory Tych of
Top, with reflector ρ, where ρ is the Tychonoff reflector, a continuous map f
between topological spaces is orthogonal to the unit interval I = [0, 1] equipped
with its usual topology if and only if its Tychonoff reflection ρ(f) is orthogonal
to I.

Let us now give some properties of a continuous map orthogonal to I (resp.,
[0, +∞[).

Proposition 2.5. Let f be a continuous map from a functionally Hausdorff
space to a topological space Y . If f is orthogonal to I, then f is one-to-one.

Proof. Let x and y be two points in X such that f(x) = f(y). If x and y are
distinct then there exists g ∈ C(X) such that g(x) = 0 and g(y) = 1 and thus
gI defined in Remark 2.2 satisfies also gI(x) = 0 and gI(y) = 1 . The mapping
g̃I in C

⋆
I(Y ) obtained by orthogonality of f to I gives a contradiction. �

Remark 2.6. By the same way as in Proposition 2.5, we can see easily that if we
consider a continuous map from a functionally Hausdorff space to a topological
space Y which is orthogonal to [0, +∞[, then it is one to one. Indeed, it is
enough to replace gI in Proposition 2.5 by |g|.

Proposition 2.7. Let f be a continuous map from a topological space X to a
completely regular space Y . If f is orthogonal to I, then, f is a dense mapping.

Proof. Assume that f(X) 6= Y and let y be in Y and not in f(X). Since Y
is completely regular, there exists a mapping h in C(Y ) such that h(y) = 0

and h(f(X)) = {1}. Then the mapping hI from C
⋆
I(X) satisfies hI(y) = 0 and

hI(f(X)) = inf{|f|, 1}(f(X)) = inf{1, 1}(f(X)) = {1}. Now if we denote by
1Y the constant map equal to 1 from Y to I, we get:

∀x ∈ X, (1Y ◦ f)(x) = (hI ◦ f)(x) = 1.

So,

1Y ◦ f = hI ◦ f.

This leads to a contradiction because f is orthogonal to I and the continuous
maps 1Y and hI are not equal. �

Remark 2.8. By the same way as in Proposition 2.7, we can see easily that
if we consider a continuous map from a topological space X to a completely
regular space Y which is orthogonal to [0, +∞[, then it is a dense mapping.
Indeed, it is enough to replace hI in Proposition 2.7 by |h|.

Proposition 2.9. Let f be a continuous map from a completely regular space
X to a topological space Y . If f is orthogonal to I, then f(X) and X are
homeomorphic.

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S. Lazaar and S. Nacib

Proof. Let f1 be the restriction of f to f(X). Using Proposition 2.5, f1 is a
continuous bijective map, so it is sufficient to show that it is an open map.

Indeed, let g−1(]0, 1]) be an element of the base of open sets, cited in Remark
2.2, where g ∈ C⋆I(X). Since f is orthogonal to I, the unique map g̃ ∈ C

⋆
I(Y )

such that g̃ ◦ f = g satisfies f1(g
−1(]0, 1])) = g̃−1(]0, 1]) ∩ f(X), which is open

in f(X). �

Remark 2.10. By the same way as in Proposition 2.9, any continuous map f
from a completely regular space X to a topological space Y which is orthogonal
to [0, +∞[, then f(X) and X are homeomorphic.

To conclude the three previous results, we can cite the following result.

Proposition 2.11. Every map f : X −→ Y in the category Tych which is
orthogonal to I (resp., [0, +∞[ ) is a one-to-one dense mapping such that X
and f(X) are homeomorphic.

Proposition 2.12. Let X be a Tychonoff space and f : X −→ I be a contin-
uous map which is orthogonal to I. Then f is an homeomorphism.

Proof. By Proposition 2.11, it is enough to see that f is a surjective map.
Suppose that f(X) 6= I and let y be in I not in f(X).

We have two cases to discuss.
First case: 0 < y < 1.
Let us denote by:

X
< = {x ∈ X : 0 ≤ f(x) < y} and X> = {x ∈ X : y < f(x) ≤ 1}.

So, one can check easily that X< and X> are a disjoint union of X. Then

f(X) = f(X<) ∪ f(X>) which implies that I = f(X) = f(X>) ∪ f(X>).

Now since f(X<) (resp., f(X>)) is closed containing [0, y[ (resp., ]y, 1] ), then

f(X<) = [0, y] (resp., f(X>) = [y, 1]).
Let consider the map g from X to I by g(X<) = {y

2
} and g(X>) = {y+1

2
}.

It is clear that g is continuous and thus by orthogonality of f to I, let g̃ be the
continuous map from I to itself such that g̃ ◦ f = g.

By density of f(X<) (resp.,f(X>)) in [0, y] (resp., [y, 1] ), consider a se-
quence (xn) (resp., (zn) ) in X

< (resp., X> ) with (f(xn)) (resp., (f(zn))) in
[0, y] (resp., [y, 1] ) converges to y. By preserving continuity under continu-

ous maps, the constant sequences (g(xn) =
y
2
) and (g(zn) =

y+1
2

) must both
converge to g̃(y) which is impossible.

Second case: y ∈ {0, 1}.
In this case f(X) ∈ {]0, 1], [0, 1[, ]0, 1[}. Without loss in generality we can

suppose that f(X) =]0, 1]. Now, consider the map g from X to I defined by
g(x) =| sin 1

f(x)
|. Clearly g is a continuous map and since f is orthogonal to I,

there exists a unique map g̃ from I to itself such that g̃ ◦ f = g. So that for any
y ∈]0, 1], g̃(y) =| sin 1

y
| which leads to a contradiction since g̃ is continuous in

0. �

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The hull orthogonal of the unit interval [0,1]

3. Hcomp ={I}⊥⊥

Before giving the main result of our paper, let us recall two important results
introduced in chapter 6 in [5].

Theorem 3.1 ([5, Theorem 6.4]). Let X be dense in T . The following state-
ments are equivalent.

(1) Every continuous mapping τ from X into any compact space Y has an
extension to a continuous mapping from T into Y .

(2) X is C⋆-embedded in T .
(3) Any two disjoint zero-sets in X have disjoint closures in T .
(4) For any two zero-sets Z1 and Z2 in X,

clT (Z1 ∩ Z2) = clT Z1 ∩ clT Z2.

(5) Every point of T is the limit of a unique z-ultrafilter on X.

Theorem 3.2 (Compactification Theorem, [5, Theorem 6.5]). Every com-
pletely regular space X has a compactification βX, with the following equivalent
properties.

(1) (Stone) Every continuous mapping τ from X into any compact space
Y has a continuous extension τ from βX into Y .

(2) (Stone-Cech) Every function f in C⋆(X) has an extension to a function
fβ in C(βX).

(3) (Cech) Any two disjoint zero-sets in X have disjoint closures in βX.
(4) For any two zero-sets Z1 and Z2 in X,

clβX(Z1 ∩ Z2) = clβXZ1 ∩ clβXZ2.

(5) Distinct z-ultrafilter on X have distinct limits in βX.

Remark 3.3. The compactification βX in Theorem 3.2 is unique, in the follow-
ing sense: if a compactification T of X satisfies anyone of the listed previous
conditions, then there exists a homeomorphism from βX onto T that leaves X
pointwise fixed.

Now, we are in a position to give our main result.

Theorem 3.4. Hcomp
⊥ = {I}⊥.

Proof. Clearly, Hcomp⊥ ⊂ {I}⊥.
Conversely, let f : X −→ T be a continuous map orthogonal to I, Y a

Hausdorff compact space and g a continuous map from X to Y .
By Remark 2.4, we may assume X and T are completely regular spaces.

Now, using Proposition 2.11, we may assume X as a dense subset of the com-
pletely regular space T and replace f by the canonical injection from X to T.
Now (2) =⇒ (1) of Theorem 3.1 applies, and thus g has a continuous extension
g̃ from T into Y . Furthermore, this extension is unique, since any two such
continuous extensions must coincides on the dense subset X of the Hausdorff
space T , and thus must be equal. �

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S. Lazaar and S. Nacib

The following corollaries are immediate.

Corollary 3.5. Hcomp = {I}⊥⊥.

Corollary 3.6. Let f be a continuous map. Then β(ρ(f)) is a homeomorphism
if and only if f is orthogonal to I. In particular, for a continuous map f between
two Tychonoff spaces, β(f) is an homeomorphism if and only if f is orthogonal
to I.

Proof. Since the family of all morphisms rendered invertible by the reflector
β ◦ ρ is exactly Hcopm⊥, an application of Theorem 3.4 gives the result. �

Let us recall the definition introduced by Echi and Lazaar in [3].

Definition 3.7 ([3, Definition 3.2]). Let X be a topological space and H a
subset of C(X). We say that H has the finite intersection property (FIP, for
short) if for each finite subset J of H we have ∩[f−1({0}, f ∈ J] 6= ∅.

Theorem 3.8. Let f : X −→ Y be a continuous map which is orthogonal to
I. Then the following statements are equivalent.

(1) For each subset H of C(Y ) satisfying the FIP, ∩[f−1({0}) : f ∈ H] 6=
∅;

(2) β(ρ(X)) = ρ(Y ).

Proof. (1) =⇒ (2) By [3, Proposition 3.6] ρ(Y ) is a completely regular compact
space. Then ρ(f) is a continuous map from the Tychonoff space ρ(X) to the
compact Tychonoff space ρ(Y ). Using the previous results, one can see that
ρ(f)(ρ(X)) is a dense subset of the compact Hausdorff space ρ(Y ) = β(ρ(Y ))
which is C⋆-embedding. Hence by the Theorem 3.2 (2) and the Remark 3.3,
β(ρ(f)(ρ(X)) = ρ(Y ). Finally, since ρ(f)(ρ(X)) and ρ(X) are homeomorphic,
(2) is satisfied.

(2) =⇒ (1) is an immediate consequence of [3, Proposition 3.6]. �

Corollary 3.9. Let f : X −→ Y be a continuous map between Tychonoff
spaces, with f⊥ I. Then the following statements are equivalent.

(1) Y is compact;
(2) β(X) = Y (up to homeomorphism).

Examples 3.10. (1) Let i :]0, 1] −→ [0, 1] be the canonical injection.
Clearly i is a dense mapping between two Tychonoff spaces and the
second space is compact. So, β(]0, 1]) 6= [0, 1] because i is not orthog-
onal to I. Indeed the continuous mapping g :]0, 1] → [0, 1] defined by

g(x) = | sin(
1

x
)| can not be extended to x = 0.

(2) Let i :]0, 1] −→]0, 1] the canonical injection. Clearly i is a dense map-
ping between two Tychonoff spaces. Since i is an isomorphism it is
orthogonal to I, but β(]0, 1]) 6=]0, 1] because ]0, 1] is not compact.

Remark 3.11. Regarding [1], the authors in [1, Proposition 4.11], proved that,
for any continuous map f : X −→ Y between two Tychonoff spaces, β(f) is

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The hull orthogonal of the unit interval [0,1]

a homeomorphism if and only if β(f(X)) = β(Y ), so in our case this result
becomes trivial because β(f) : β(X) −→ β(Y ) is a homeomorphism if and
only if f is orthogonal to I and in this situation, by Proposition 2.9, X is
homoeomorphic to f(X) and consequently β(f(X)) is homeomorphic to β(X).
Finally, β(f(X)) = β(Y ).

To finish this paper, we shield some light on the hull orthogonal of a given
topological space. By [6], Hewitt = {R}⊥⊥ and it is clear that any homeomor-
phic topological space to R satisfies also this property. The following example
shows that the topological space [0, +∞[, which is not homeomorphic to R,
satisfies also Hewitt = {[0, +∞[}⊥⊥.

Proposition 3.12. Hewitt = {[0, +∞[}⊥⊥.

Proof. Since {R}⊥ = Hewitt
⊥

and {[0, +∞[} ⊂ Hewitt, then {R}⊥ ⊂
{[0, +∞[}⊥.

Conversely, let f : X −→ Y be a continuous map which is orthogonal to
[0, +∞[ and let us show that it is orthogonal to R. By Proposition 2.3, we can
suppose that X and Y are Tychonoff spaces. Now according to Proposition
2.11, we can suppose that X is a dense subset of a Tychonoff space Y and f is
the canonical injection from X to Y .

For this, let g be a continuous map from X to R. Then g+ = max(g, 0)
( resp., −g− = − min{g, 0} ) is a continuous map from X to [0, +∞[. By

orthogonality of f to [0, +∞[, there exists a continuous map g̃+ (resp.,−̃g−)

from Y to [0, +∞[ such that g̃+ ◦ f = g+ (resp., (−̃g−) ◦ f = (−g−) ). Hence

(g̃+ − (−̃g−)) ◦ f = (g̃+ ◦ f) − ((−̃g−) ◦ f) = (g+) − (−g−) = g+ + g− = g.

So the existence of a continuous map g̃ = g̃+ − −̃g− from Y to R such that
g̃ ◦ g = f. The uniqueness of a such function follows immediately from the
density of X in Y and the fact that R is Hausdorff. �

Acknowledgements. The authors thank the referee for his/her comments,
corrections and suggestions improving both the presentation and the math-
ematical content of this paper. Lazaar would like to thank The laboratory
of research LATAO (Faculty of sciences of Tunis, University Tunis El Manar,
Tunisia) for its support (LR11ES12).

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S. Lazaar and S. Nacib

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